Observe that the set \(\cup _{t< 0} \sqrt{-t} \bar{\Omega } \times \lbrace t\rbrace \)  is invariant under parabolic dilation around the space-time origin. We thus conclude that for all \(\lambda \in [0,\infty )\)  and \(t < \min \lbrace 0,\lambda ^2 t_0\rbrace \) ,
\(\sqrt{\lambda ^2t_0 -t} \Sigma +\lambda \mathbf {x}_0 \subset \sqrt{-t} \, \bar{\Omega }\) 
In particular, taking \(t=-1\)  and \(\lambda \ge 0\)  small, we have that
\(\lambda \mapsto \Sigma _\lambda : = \sqrt{1 + \lambda ^2t_0} \Sigma +\lambda \mathbf {x}_0 \subset \bar{\Omega }\) 
is a 1-parameter family of hypersurfaces with \(\Sigma _0 = \Sigma = \partial \Omega \) . The normal speed at \(\lambda =0\)  is \(\mathbf {x}_0 \cdot \nu _\Sigma \ge 0\)  (where \(\nu _\Sigma \)  is the unit normal pointing into \(\Omega \) ). Because
\(\Delta _\Sigma (\mathbf {x}_0 \cdot \nu _\Sigma ) - \tfrac{1}{2} \mathbf {x}\cdot \nabla _\Sigma (\mathbf {x}_0 \cdot \nu _\Sigma ) + |A_\Sigma |^2 (\mathbf {x}_0\cdot \nu _\Sigma ) = 0\) 
(cf. [1]}), the maximum principle implies that either \(\mathbf {x}_0\cdot \nu _\Sigma >0\)  along \(\Sigma \)  or \(\mathbf {x}_0\cdot \nu _\Sigma = 0\)  along \(\Sigma \) . (Note that \(\Sigma \)  is connected thanks to the Frankel property of shrinkers, cf. [2]}.)
In the first case (i.e., \(\mathbf {x}_0\cdot \nu _\Sigma >0\) ), each component of \(\Sigma \)  is a graph over the \(\mathbf {x}_0^\perp \) -hyperplane. By [3]} (cf. [4]}), each component of \(\Sigma \)  must be a hyperplane, so there is only one component and \(\Sigma \)  is a flat hyperplane. This contradicts the assumption that \(\Sigma \in \mathcal {S}_n^*\)  (the set of non-flat shrinkers).
In the second case (i.e., \(\mathbf {x}_0\cdot \nu _\Sigma = 0\) ), we see that \(\mathbf {x}_0\in T_p\Sigma \)  for all \(p\in \Sigma \) . In particular, if \(\mathbf {x}_0\ne 0\) , then \(\Sigma \)  splits a line in the \(\mathbf {x}_0\) -direction. It thus remains to consider the situation in which \(\mathbf {x}_0=\mathbf {0}\) . If this is the case, then it must hold that \(t_0\ne 0\)  and we have
\(\tilde{\Sigma }_\mu := (1+\mu t_0) \Sigma \subset \bar{\Omega }.\) 
for \(\mu \ge 0\)  sufficiently small. The normal speed at \(\mu =0\)  is \(t_0 \mathbf {x}\cdot \nu _\Sigma \ge 0\) . Using the shrinker equation, we thus find that \(t_0 H_\Sigma \ge 0\) . Since \(t_0\ne 0\) , we can assume that \(H_\Sigma \ge 0\) . Thus, up to a rotation, \(\Sigma = \mathbb {S}^{k}(\sqrt{2k})\times \mathbb {R}^{n-k}\)  for \(k=1,\dots ,n\)  by [1]}. This completes the proof.
